Chi-Square Homogeneity or Independence (Test)
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AP Statistics › Chi-Square Homogeneity or Independence (Test)
A researcher wants to compare preferred news source across two age groups. Independent random samples of adults ages 18–34 and ages 35+ were surveyed, and each person selected one primary news source (TV, Online, Print). A chi-square test of homogeneity was performed at $\alpha=0.05$. The observed counts and p-value are shown.
What conclusion is appropriate?
Because the p-value is greater than 0.05, the sample proves the population distributions are identical.
Because the p-value is greater than 0.05, there is not convincing evidence that the distribution of news source preference differs between the two age groups.
Because the p-value is greater than 0.05, there is convincing evidence that the distribution of news source preference differs between the two age groups.
Because the p-value is less than 0.05, there is not convincing evidence that the distributions differ.
Because the p-value is greater than 0.05, age group causes people to choose the same news source.
Explanation
This chi-square test of homogeneity compares news source preferences between two age groups. With a p-value greater than 0.05, we fail to reject the null hypothesis that both age groups have the same distribution of news source preferences, meaning there is not convincing evidence that the distributions differ. Choice A incorrectly concludes there is evidence of differences. Choice C wrongly introduces causation. Choice D has the wrong p-value comparison. Choice E overstates by claiming to prove identical distributions. Important distinction: not finding evidence of differences doesn't prove the groups are identical—it means any differences aren't statistically significant.
A university randomly sampled students and recorded their class standing (First-year, Sophomore, Junior, Senior) and whether they participate in a campus club (Yes/No). The two-way table is shown. A chi-square test of independence gave $p$-value $=0.096$. What conclusion is appropriate at the 5% significance level?
Because $p=0.096$, club participation causes students to become seniors.
Because $p<0.05$, there is convincing evidence that class standing and club participation are associated.
Because $p=0.096$, there is not convincing evidence of an association between class standing and club participation at the 5% level.
Because $p=0.096$, the sample shows that exactly 9.6% of students participate in clubs.
Because $p=0.096$, we can conclude class standing and club participation are independent in the population.
Explanation
This chi-square test of independence examines association between class standing and club participation. The p-value of 0.096 is greater than the significance level of 0.05, so we fail to reject the null hypothesis. Choice B correctly states there is not convincing evidence of association at the 5% level. Choice A incorrectly claims p < 0.05. Choice C overstates—failing to find evidence of association doesn't prove independence. Choice D absurdly suggests reverse causation. Choice E misinterprets the p-value as a percentage of participation. Important: p-values are probabilities of getting results at least as extreme as observed if the null hypothesis is true, not population percentages.
A researcher randomly assigned 180 volunteers to one of three diets (Diet A, Diet B, Diet C) and recorded whether each volunteer lost weight after 8 weeks (Yes/No). The two-way table is shown. A chi-square test of homogeneity for the distribution of weight-loss outcomes across diets produced $p$-value $=0.018$. What conclusion is appropriate at the 5% significance level?
Because $p<0.05$, the diets and weight loss are independent.
Because $p<0.05$, the sample proves Diet A is the best diet for all people.
Because $p>0.05$, there is not convincing evidence of differences in weight-loss outcomes among the diets.
Because $p<0.05$, volunteers’ personal motivation caused the differences, not diet assignment.
Because $p<0.05$, there is convincing evidence that the weight-loss outcome distribution is not the same for all three diets.
Explanation
This is a chi-square test of homogeneity comparing weight-loss distributions across three diets. The p-value of 0.018 is less than 0.05, so we reject the null hypothesis that all diets have the same distribution of outcomes. Choice A correctly concludes there is evidence the distributions differ. Choice B incorrectly claims independence when we've found differences. Choice C overstates by claiming one diet is best. Choice D misreads p as greater than 0.05. Choice E incorrectly dismisses the randomized design. Since this was a randomized experiment (not observational), we could potentially make causal claims, though the correct answer focuses on the statistical conclusion.
A counselor examines whether participation in an after-school program is associated with whether students report feeling stressed. A random sample of students is surveyed and classified as Program (Yes/No) and Stressed (Yes/No). The results are shown below. A chi-square test of independence yields $p=0.94$. What conclusion is appropriate at the $\alpha=0.05$ level?
Two-way table (counts):
| Program participation | Stressed: Yes | Stressed: No |
|---|---|---|
| Yes | 48 | 52 |
| No | 47 | 53 |
There is not convincing evidence of an association between program participation and stress among students.
There is convincing evidence that program participation reduces stress.
There is convincing evidence of an association between program participation and stress.
We can conclude that 94% of students are not stressed.
Because $p=0.94$, the null hypothesis must be false.
Explanation
This question involves the chi-square test of independence to investigate association between program participation and stress. With p=0.94 much greater than alpha=0.05, we fail to reject the null, finding no convincing evidence of association. Choice A is a distractor implying causation, that participation reduces stress, but no association was detected. Association means variables are linked statistically, whereas causation requires demonstrating one affects the other, impossible from this survey data. The nearly equal proportions in the table yield the high p-value, indicating chance variation. This example shows how balanced data can lead to non-significant results.
An ecologist took independent random samples of trees from two parks (Park A and Park B) and classified each tree as Healthy, Diseased, or Dead. The results are shown in the two-way table. A chi-square test of homogeneity comparing the distribution of tree condition across the two parks produced $p$-value $=0.007$. What conclusion is appropriate at the 1% significance level?
Because $p=0.007$, the probability a tree is diseased is 0.007 in each park.
Because $p<0.01$, the trees in the two parks must have been sampled without randomness.
Because $p<0.01$, there is convincing evidence that the distribution of tree condition differs between Park A and Park B.
Because $p<0.01$, the park location causes individual trees to become diseased.
Because $p>0.01$, there is not convincing evidence that the distribution of tree condition differs between the parks.
Explanation
This chi-square test of homogeneity compares tree condition distributions between two parks. The p-value of 0.007 is less than the significance level of 0.01, so we reject the null hypothesis and conclude the distributions differ. Choice A correctly states there is convincing evidence of different distributions. Choice B incorrectly implies the park location causes disease in individual trees. Choice C misreads 0.007 as greater than 0.01. Choice D misinterprets the p-value as a disease probability. Choice E wrongly questions the sampling method. The small p-value indicates the observed differences in tree conditions between parks are unlikely due to chance alone.
A city council wants to know whether support for a new recycling ordinance is related to neighborhood. A random sample of residents was taken, and each resident was classified by neighborhood (North, South, East) and response (Support, Oppose). A chi-square test of independence was conducted at $\alpha=0.05$. The observed counts and p-value are shown.
What conclusion is appropriate?
Because the p-value is greater than 0.05, there is convincing evidence of an association between neighborhood and support.
Because the p-value is less than 0.05, the sample proves the exact proportion supporting in each neighborhood.
Because the p-value is less than 0.05, the results show no association between neighborhood and support.
Because the p-value is less than 0.05, there is convincing evidence that support for the ordinance is associated with neighborhood in the city.
Because the p-value is less than 0.05, living in a particular neighborhood causes a resident to support the ordinance.
Explanation
This chi-square test of independence examines whether support for a recycling ordinance is related to neighborhood. With a p-value less than 0.05, we reject the null hypothesis of independence and conclude there is convincing evidence of an association between neighborhood and support for the ordinance. Choice B incorrectly claims causation—living in a neighborhood doesn't necessarily cause support; other factors may be involved. Choice C contradicts the finding by claiming no association. Choice D has the wrong p-value comparison. Choice E overstates by claiming to prove exact proportions. Association indicates a relationship exists but doesn't explain why or establish cause-and-effect.
A marketing team wants to know whether device type is related to whether a customer completes an online purchase. From a random sample of site visits, each visit was classified by device (Phone, Tablet, Computer) and outcome (Purchase, No purchase). A chi-square test of independence was performed at the $\alpha=0.05$ level. The observed counts and p-value are shown.
What conclusion is appropriate?
Because the p-value is less than 0.05, there is not convincing evidence of an association between device type and purchase outcome.
Because the p-value is greater than 0.05, there is not convincing evidence of an association between device type and purchase outcome for site visits.
Because the p-value is greater than 0.05, the sample shows that exactly the same percentage purchase on each device in the population.
Because the p-value is greater than 0.05, device type and purchase outcome are proven to be independent in all circumstances.
Because the p-value is greater than 0.05, device type causes purchase outcome to be the same across devices.
Explanation
This chi-square test of independence examines whether device type and purchase outcome are associated. With a p-value greater than 0.05, we fail to reject the null hypothesis of independence, meaning there is not convincing evidence of an association between the variables. Choice B overstates by claiming to prove independence in all circumstances. Choice C incorrectly introduces causation. Choice D has the wrong p-value comparison. Choice E misinterprets the result as proving exact percentages. Important concept: failing to reject the null hypothesis doesn't prove independence—it simply means we lack sufficient evidence to claim association.
A company wants to know whether preferred work arrangement is associated with department. A random sample of employees was selected, and each employee reported a preferred arrangement (Remote, Hybrid, In-office) and department (Engineering, Sales). A chi-square test of independence was conducted at $\alpha=0.05$. The observed counts and p-value are shown.
What conclusion is appropriate?
Because the p-value is greater than 0.05, department causes employees to have the same preferences.
Because the p-value is greater than 0.05, there is convincing evidence that preference and department are associated.
Because the p-value is greater than 0.05, the sample proves the population preferences are identical across departments.
Because the p-value is greater than 0.05, there is not convincing evidence of an association between department and preferred work arrangement among employees.
Because the p-value is less than 0.05, there is not convincing evidence of an association.
Explanation
This chi-square test of independence examines whether work arrangement preference is associated with department. With a p-value greater than 0.05, we fail to reject the null hypothesis of independence, meaning there is not convincing evidence of an association between department and preferred work arrangement. Choice A incorrectly concludes there is evidence of association. Choice C wrongly introduces causation. Choice D has the wrong p-value comparison. Choice E overstates by claiming to prove identical preferences. Key concept: failing to find evidence of association doesn't mean the variables are definitely independent—it means we lack sufficient evidence to claim they're related at the 0.05 significance level.
A botanist tests whether three fertilizers lead to different distributions of plant growth categories. She randomly assigns plants to Fertilizer A, B, or C, then classifies growth after 8 weeks as Low, Medium, or High. The results are shown below. A chi-square test of homogeneity is performed and gives $p<0.001$. What conclusion is appropriate?
Two-way table (counts):
| Fertilizer | Low | Medium | High |
|---|---|---|---|
| A | 25 | 40 | 15 |
| B | 10 | 35 | 35 |
| C | 30 | 30 | 20 |
We can conclude Fertilizer B causes high growth for all plants.
There is convincing evidence that fertilizer choice and growth category are independent.
We should fail to reject $H_0$ because the sample size is the same in each fertilizer group.
There is not convincing evidence of differences because $p<0.001$ is too small to be reliable.
There is convincing evidence that the distribution of growth categories differs among the fertilizers.
Explanation
This question tests the chi-square test of homogeneity, comparing distributions of growth categories across fertilizer groups. The p-value less than 0.001 is well below a typical alpha=0.05, so we reject the null hypothesis and conclude there is convincing evidence of differences in distributions. Choice D is a distractor that overstates causation, claiming Fertilizer B causes high growth for all plants, but the test only shows differing distributions. Association in this context means the growth outcomes vary by fertilizer, but causation would need confirmation that the fertilizer directly affects growth, perhaps through controlled trials. The table reveals Fertilizer B has more high growth, contributing to the significant result. Equal sample sizes per group support the test's assumptions, but the conclusion hinges on the p-value.
A researcher wants to determine whether political affiliation is associated with preferred news source among adults in a state. A random sample of adults is surveyed and classified by affiliation (Democrat, Republican, Independent) and preferred news source (TV, Online, Print). The results are shown below. A chi-square test of independence yields $p=0.072$. What conclusion is appropriate at the $\alpha=0.05$ level?
Two-way table (counts):
| Affiliation | TV | Online | |
|---|---|---|---|
| Democrat | 60 | 85 | 15 |
| Republican | 75 | 70 | 20 |
| Independent | 55 | 80 | 20 |
Because $p=0.072$, we should reject $H_0$ at $\alpha=0.05$.
There is not convincing evidence of an association between preferred news source and political affiliation in the state.
The results show that political affiliation causes people to choose different news sources.
We can conclude that 7.2% of adults prefer TV news.
There is convincing evidence that preferred news source and political affiliation are associated in the state.
Explanation
This question involves the chi-square test of independence to check for an association between political affiliation and preferred news source. Since the p-value of 0.072 exceeds alpha=0.05, we fail to reject the null hypothesis, indicating no convincing evidence of an association. Choice D serves as a distractor by incorrectly suggesting causation, that affiliation causes news source preferences, even though no association was found. Association means the variables covary statistically, whereas causation implies a directional influence, which this observational study cannot prove. The table's counts show similar patterns across affiliations, explaining the higher p-value. This outcome suggests any observed differences could be due to random sampling variation.